Problem: In the right triangle shown, $\angle B = 60^\circ$ and $AB = 6\sqrt{3}$. How long is $AC$ ? $A$ $C$ $B$ $x$ $6\sqrt{3}$
Solution: We know the length of the hypotenuse of this triangle, and want to find the length of a leg. What mathematical relationship is there between a right triangle's legs and its hypotenuse? We can use either sine (opposite leg over hypotenuse) or cosine (adjacent leg over hypotenuse). This is a 30-60-90 triangle, so we know what the values of sine and cosine are at each angle of the triangle. Let's try using sine: $A$ $C$ $B$ $x$ $6\sqrt{3}$ ${60}^{\circ}$ Sine is opposite over hypotenuse (SOH CAH TOA), so $\sin {60}^{\circ} = \dfrac{x}{6\sqrt{3}}$ . We also know that $\sin{60}^{\circ} = \dfrac{\sqrt{3}}{2}$ Solving for $x$ , we get $ x = 6\sqrt{3} \cdot \sin{60}^{\circ}$ $ x = 6\sqrt{3} \cdot \dfrac{\sqrt{3}}{2}$ So, $x = 9$.